Is there any non-trivial relationship between kernels & kernel pairs? Kernels are inspired by group theory, and kernel pairs by a similar concept in monoids where kernels aren't sufficient to capture the information necessary for the first isomorphism theorem.
When a category is enriched over groups it seems to me that the two concepts can be shown to be the same. 
Is there some non-trivial relationship outside of enriched category theory that throws light on how the two are related?
This idea didn't work.
 A: The category of groups $\mathsf{Grp}$ is not monoidal, therefore it doesn't really make sense to talk about $\mathsf{Grp}$-enriched categories. For $\mathsf{Ab}$-enriched categories, i.e. linear categories, it is correct that the kernel pair is isomorphic to the kernel. More precisely, let $f : X \to Y$ be a morphism, admitting a kernel $K \hookrightarrow X$ and a kernel pair $X \times_Y X$. Then $K \hookrightarrow X$ and $0 : K \to X$ induce a morphism $K \to X \times_Y X$. Conversely, $p_1-p_2 : X \times_Y X \to X$ factors through a morphism $X \times_Y X \to K$. These two morphisms inverse to each other. This can be checked directly, or may be reduced to the well-known case of abelian groups using the the enriched Yoneda Lemma.
For arbitrary categories with zero morphisms (so that we can define kernels), there is a canonical map $K \times K \to X \times_Y X$, as well as a diagonal map $K \to X \times_Y X$. But they are almost never isomorphisms.
In general, there is no isomorphism between the kernel and the kernel pair, or something derived from that. You can alreay see this for algebraic categories. For example, look at pointed sets. The kernel of a map of pointed sets $f : (X,x_0) \to (Y,y_0)$ is $\{x \in X : f(x)=y_0\}$ (together with its inclusion to $X$ and the base point $x_0$), but the kernel pair is $\{(x,x') \in X \times X : f(x)=f(x')\}$ (together with the projections and the base point $(x_0,x_0)$). Besides the diagonal $x \mapsto (x,x)$, there is no map between them (depending naturally on $f$).
